Sequence of Powers of Number less than One/Necessary Condition/Proof 4

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Theorem

Let $x \in \R$ be such that $\size{x} < 1$.

Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = x^n$.


Then $\sequence {x_n}$ is a null sequence.


Proof

Define:

$\ds L = \inf_{n \mathop \in \N} \size x^n$

By the Continuum Property, such an $L$ exists in $\R$.


Clearly, $L \ge 0$.

Aiming for a contradiction, suppose $L > 0$.

Then, by the definition of the infimum, we can choose $n \in \N$ such that $\size x^n < L \size x^{-1}$.

But then $\size x^{n + 1} < L$, which contradicts the definition of $L$.

Therefore, $L = 0$.


Let $\epsilon \in \R_{>0}$ be a strictly positive real number.

By the definition of the infimum, there exists an $N \in \N$ such that $\size x^N < \epsilon$.

It follows that:

$\forall n \in \N: n \ge N \implies \size {x^n} = \size x^n \le \size x^N < \epsilon$

where Absolute Value Function is Completely Multiplicative is applied.


Hence the result, by the definition of a limit.

$\blacksquare$